Coin Toss and Kelly Criterion I read a hypothetical probability experiment in https://en.wikipedia.org/wiki/Kelly_criterion under the section "Optimal betting example" and I have two questions.
I conclude the following parameters from the provided hypothetical experiment:
Initial Bankroll = $25
Probability of Head = 60%
Total Bets = 300
Maximum Prize = $250
Question 1: How does Wiki article provide an answer of "...betting only 12% of the pot on each toss...(a 95% probability of reaching the cap and an average payout of $242.03)"?
Could somebody explain how to get these results of betting 12% will have 95% probability of reaching the average payout of $242.03?
Question 2: What is one standard deviation value and the lowest 5% distribution in the above hypothetical experiment?
I have attached my understanding in google sheet: https://docs.google.com/spreadsheets/d/1-2ii12JxXVeSNtBKl3sRyzpZS5i8uhAq/edit#gid=1151989864
Any help in form of written explanation along with recursive or non-recursive formula (no VBA) in the Excel file would be very appreciated!
 A: It is worth reading the original article, if only for the description of how real people behaved irrationally.  In what follows, I have not rounded to the nearest cent, and it is unlikely to make a substantial difference.
If your strategy is to be a fraction $f$ of your bankroll each time, than after $a$ successes and $b$ failures you would have a bankroll of $$25(1+f)^a(1-f)^b$$ at least so long as this does not exceed $250$; if it does exceed $250$ then you stop with a result of $250$.  Otherwise you stop when $a+b=300$ with however much you have.
For example with $f=0.12$ and $a=63$ and $b=37$, you get $25(1+f)^a(1-f)^b = 278.3152$ so you would stop with $250$;  with $f=0.12$ and $a=160$ and $b=140$, you get $25(1+f)^a(1-f)^b = 31.65175$ so you would stop with that since $a+b=300$ and it is less than $250$.
To work out the probabilities let $p(a,b)$ be the probability of reaching $a$ successes and $b$ failures without having previously stopped.  You have $$p(a,b)=0.6p(a-1,b)\, I_{\text{not stopped by }a-1,b}+0.4p(a,b-1)\, I_{\text{not stopped by }a,b-1}$$ starting at $p(0,0)=1$.
For example with $f=0.12$ and $a=63$ and $b=37$ you get $p(63,37)=0.01022$ while with $f=0.12$ and $a=160$ and $b=140$ you get $p(160,140)=0.00267$.  This are smaller than the corresponding binomial probabilities of $0.06820$ and $0.00299$ because of the possibility of having stopped earlier.
To find the probability of reaching the cap of $250$, you just add up all the probabilities of stopping at the cap, which for $f=0.12$ range from $p(21,0)$ to $p(169,131)$ and comes to $0.95021$.  Similarly the expected outcome is just the sum of the stopping amounts multiplied by their probabilities in the usual way, which for $f=0.12$ comes to $ 242.0325$. For $f=0.2$ the probability of hitting the cap is  $0.93823$ and the expected gain is $237.3587$.
